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The current $I$ through a rod of a certain metallic oxide is given by $I = 0.2{V^{5/2}}$, where $V$ is the potential difference across it. The rod is connected in series with a resistance to a $6V$ battery of negligible internal resistance. What value should the series resistance have so that the power dissipated in the rod is twice that dissipated in the resistance?

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Last updated date: 14th Jun 2024
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Answer
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Hint The question is asking us about the value of resistance so that the power dissipated in the rod is twice of that dissipated in the resistance, so, you simply need to find the value of resistance of the rod using the given relation of current and voltage and then, find the value of the resistance using the value of the resistance of rod and the relation of powers dissipated in both parts. Do remember that the current flowing through both the parts will be exactly the same as they are in series.

Complete step by step answer:
We will proceed with the solution exactly as explained in the hint section of the solution to the asked question. First, we will use the given relation between the current and voltage to find the value of the resistance of the rod, then we will use the value of the resistance of the metallic rod to find the value of the resistance in series with the help of the relation given between the powers dissipated in both parts.
Let us first have a look at what is given to us in the question for the metallic rod section:
$I = 0.2{V^{5/2}}$
Where $I$ is the current flowing through the metallic rod and,
$V$ is the potential difference across the metallic rod
We can also write it as:
\[I = \left( {\dfrac{1}{5}{V^{3/2}}} \right)V\] … (i)
If we represent the resistance of the metallic rod as $r$, we can write:
$r = \dfrac{V}{I}$
If we do a transpose in the equation (i), we can write:
\[
  I = \left( {\dfrac{1}{5}{V^{3/2}}} \right)V \\
   \Rightarrow \dfrac{V}{I} = \dfrac{5}{{{V^{3/2}}}} \\
   \Rightarrow r = \dfrac{5}{{{V^{3/2}}}} \\
 \]
Now, the question has told us that the power dissipated in the rod is twice of that of the power dissipated in the resistance in series, this can be mathematically represented as:
${P_M} = 2{P_R}$
Where, ${P_M}$ is the power dissipated in the metallic rod whilst,
${P_R}$ is the power dissipated in the series resistance part.
Let us assume the resistance in the series has a value of $R$
Since both the sections are in series, the current flowing through them will be the same, $I$
We can write both the powers as:
$
  {P_M} = {I^2}r \\
  {P_R} = {I^2}R \\
 $
Substituting the values in the relation, we get:
$
  {I^2}r = 2{I^2}R \\
   \Rightarrow r = 2R \\
 $
Putting in the value of \[r = \dfrac{5}{{{V^{3/2}}}}\] we get:
$R = \dfrac{5}{{2{V^{3/2}}}}$
Since both the resistances are in series, the net voltage in the circuit of $6V$ will be divided as:
$V = \dfrac{r}{{r + R}}6$
Substituting $r = 2R$, we get:
$V = \dfrac{{12R}}{{3R}} = 4V$
The value of $R$ then becomes:
$
  R = \dfrac{5}{{2{V^{3/2}}}} \\
   \Rightarrow R = \dfrac{5}{{2{{\left( 4 \right)}^{3/2}}}} \\
   \Rightarrow R = \dfrac{5}{{16}} \\
   \Rightarrow R = 0.3125\Omega \\
 $

Hence, we found out the value of the series resistance to be $R = 0.3125\Omega $

Note A common mistake that many students make is missing out on the fact that current flowing through both the sections will be the same since they are in series connection and the manipulation of finding out the potential difference across the metallic rod.